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3 years ago

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An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductan

ce of L = 0.2 henrys, a resistor of R = 5 ohms and a capacitor of capcitance C = 0.043 farads. What is the amplitude of the current I?

1 answer:

german3 years ago

4 0

Answer:

Explanation:

we have given E(t)=120 sin(12t)

R=5 ohm

L=0.2 H

ω=12 ( from expression of E)

$X_L=0.2\times 12=2.4$ ohm

$X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm$

$Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}$

$Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}$

=5.021 ohm

so amplitude of current = $\frac{v}{z}=\frac{120}{5.021}=23.89$

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2 years ago

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